Error Message:
Exception in thread "main" java.lang.NumberFormatException: For input string: "Ace of Clubs"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:580)
at java.lang.Integer.parseInt(Integer.java:615)
at set07102.Cards.main(Cards.java:68)
C:UsersqasimAppDataLocalNetBeansCache8.1executor-snippetsrun.xml:53: Java returned: 1
means:
There was an error. We try to give you as much information as possible
It was an Exception in main thread. It's called NumberFormatException and has occurred for input "Ace of Clubs".
at line 65th of NumberFormatException.java which is a constructor,
which was invoked from Integer.parseInt() which is in file Integer.java in line 580,
which was invoked from Integer.parseInt() which is in file Integer.java in line 615,
which was invoked from method main in file Cards.java in line 68.
It has resulted in exit code 1
In other words, you tried to parse "Ace of Clubs" to an int what Java can’t do with method Integer.parseInt. Java has provided beautiful stacktrace which tells you exactly what the problem is. The tool you’re looking for is debugger and using breakpoints will allow you to inspect the state of you application at the chosen moment.
The solution might be the following logic in case you want to use parsing:
if (cards[index].startsWith("Ace"))
value = 1;
else if (cards[index].startsWith("King"))
value = 12;
else if (cards[index].startsWith("Queen"))
value = 11;
...
else {
try {
Integer.parseInt(string.substring(0, cards[index].indexOf(" ")));
} catch (NumberFormatException e){
//something went wrong
}
}
What is an Exception in Java?
An exception is an event, which occurs during the execution of a
program, that disrupts the normal flow of the program’s instructions.
-Documentation
Constructors and usage in Integer#parseInt
static NumberFormatException forInputString(String s) {
return new NumberFormatException("For input string: "" + s + """);
}
public NumberFormatException (String s) {
super (s);
}
They are important for understanding how to read the stacktrace. Look how the NumberFormatException is thrown from Integer#parseInt:
if (s == null) {
throw new NumberFormatException("null");
}
or later if the format of the input String s is not parsable:
throw NumberFormatException.forInputString(s);
What is a NumberFormatException?
Thrown to indicate that the application has attempted to convert a string to one of the numeric types, but that the string does not have the appropriate format.
-Documentation
NumberFormatException extends IllegalArgumentException. It tells us that it’s more specialized IllegalArgumentException. Indeed, it’s used for highlighting that although, the argument type was correct (String) the content of the String wasn’t numeric (a,b,c,d,e,f are considered digits in HEX and are legal when needed).
How do I fix it?
Well, don’t fix the fact that it’s thrown. It’s good that it’s thrown. There are some things you need to consider:
- Can I read the stacktrace?
- Is the
Stringwhich causes anExceptionanull? - Does it look like a number?
- Is it ‘my string’ or user’s input?
- to be continued
Ad. 1.
The first line of a message is an information that the Exception occurred and the input String which caused the problem. The String always follows : and is quoted ("some text"). Then you become interested in reading the stacktrace from the end, as the first few lines are usually NumberFormatException‘s constructor, parsing method etc. Then at the end, there is your method in which you made a bug. It will be pointed out in which file it was called and in which method. Even a line will be attached. You’ll see. The example of how to read the stacktrace is above.
Ad. 2.
When you see, that instead of "For input string:" and the input, there is a null (not "null") it means, that you tried to pass the null reference to a number. If you actually want to treat is as 0 or any other number, you might be interested in my another post on StackOverflow. It’s available here.
The description of solving unexpected nulls is well described on StackOverflow thread What is a NullPointerException and how can I fix it?.
Ad. 3.
If the String that follows the : and is quoted looks like a number in your opinion, there might be a character which your system don’t decode or an unseen white space. Obviously " 6" can’t be parsed as well as "123 " can’t. It’s because of the spaces. But it can occure, that the String will look like "6" but actually it’s length will be larger than the number of digits you can see.
In this case I suggest using the debugger or at least System.out.println and print the length of the String you’re trying to parse. If it shows more than the number of digits, try passing stringToParse.trim() to the parsing method. If it won’t work, copy the whole string after the : and decode it using online decoder. It’ll give you codes of all characters.
There is also one case which I have found recently on StackOverflow, that you might see, that the input looks like a number e.g. "1.86" and it only contains those 4 characters but the error still exists. Remember, one can only parse integers with #Integer#parseInt#. For parsing decimal numbers, one should use Double#parseDouble.
Another situation is, when the number has many digits. It might be, that it’s too large or too small to fit int or long. You might want to try new BigDecimal(<str>).
Ad. 4.
Finally we come to the place in which we agree, that we can’t avoid situations when it’s user typing «abc» as a numeric string. Why? Because he can. In a lucky case, it’s because he’s a tester or simply a geek. In a bad case it’s the attacker.
What can I do now? Well, Java gives us try-catch you can do the following:
try {
i = Integer.parseInt(myString);
} catch (NumberFormatException e) {
e.printStackTrace();
//somehow workout the issue with an improper input. It's up to your business logic.
}
The NumberFormatException is an unchecked exception in Java that occurs when an attempt is made to convert a string with an incorrect format to a numeric value. Therefore, this exception is thrown when it is not possible to convert a string to a numeric type (e.g. int, float). For example, this exception occurs if a string is attempted to be parsed to an integer but the string contains a boolean value.
Since the NumberFormatException is an unchecked exception, it does not need to be declared in the throws clause of a method or constructor. It can be handled in code using a try-catch block.
What Causes NumberFormatException
There can be various cases related to improper string format for conversion to numeric values. Some of them are:
Null input string
Integer.parseInt(null);Empty input string
Integer.parseInt("");Input string with leading/trailing whitespaces
Integer myInt = new Integer(" 123 ");Input string with inappropriate symbols
Float.parseFloat("1,234");Input string with non-numeric data
Integer.parseInt("Twenty Two");Alphanumeric input string
Integer.parseInt("Twenty 2");Input string exceeding the range of the target data type
Integer.parseInt("12345678901");Mismatch of data type between input string and the target data type
Integer.parseInt("12.34");NumberFormatException Example
Here is an example of a NumberFormatException thrown when attempting to convert an alphanumeric string to an integer:
public class NumberFormatExceptionExample {
public static void main(String args[]) {
int a = Integer.parseInt("1a");
System.out.println(a);
}
}In this example, a string containing both numbers and characters is attempted to be parsed to an integer, leading to a NumberFormatException:
Exception in thread "main" java.lang.NumberFormatException: For input string: "1a"
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:68)
at java.base/java.lang.Integer.parseInt(Integer.java:652)
at java.base/java.lang.Integer.parseInt(Integer.java:770)
at NumberFormatExceptionExample.main(NumberFormatExceptionExample.java:3)Such operations should be avoided where possible by paying attention to detail and making sure strings attempted to be parsed to numeric values are appropriate and legal.
How to Handle NumberFormatException
The NumberFormatException is an exception in Java, and therefore can be handled using try-catch blocks using the following steps:
- Surround the statements that can throw an
NumberFormatExceptionin try-catch blocks - Catch the
NumberFormatException - Depending on the requirements of the application, take necessary action. For example, log the exception with an appropriate message.
The code in the earlier example can be updated with the above steps:
public class NumberFormatExceptionExample {
public static void main(String args[]) {
try {
int a = Integer.parseInt("1a");
System.out.println(a);
} catch (NumberFormatException nfe) {
System.out.println("NumberFormat Exception: invalid input string");
}
System.out.println("Continuing execution...");
}
}Surrounding the code in try-catch blocks like the above allows the program to continue execution after the exception is encountered:
NumberFormat Exception: invalid input string
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The NumberFormatException is one of the most common errors in Java applications, along with NullPointerException. This error comes when you try to convert a String into numeric data types e.g., int, float, double, long, short, char, or byte. The data type conversion methods like Integer.parseInt(), Float.parseFloat(), Double.parseDoulbe(), and Long.parseLong() throws NumberFormatException to signal that input String is not valid numeric value.
Even though the root cause is always something that cannot be converted into a number, there are many reasons and inputs due to which NumberFormatException occurs in Java applications.
Most of the time, I have faced this error while converting a String to int or Integer in Java, but there are other scenarios as well when this error occurs. In this article, I am sharing 10 of the most common reasons for java.lang.NumberFormatException in Java programs.
Having knowledge of these causes will help you to understand, analyze and solve this error in your Java application. In each case, you’ll see an error message, and then I’ll explain the reason behind it.
It’s difficult to cover all scenarios on which JVM throws this error, but I have tried to capture as many as possible here. If you find any other reasons or faced this error in your Java project due to anything mentioned below, then please share with us via comments.
10 common reasons for NumberFormatException
Here are 10 reasons which I have seen causing NumberFormatException in Java application. If you have any reason which is not listed here, but you have seen them causing this error, feel free to add it.
1. Null Input Value
Since String is an object, it’s possible that it could be null, and if you pass a null String to a method like parseInt(), it will throw NumberFormatException because null is not numeric.
Integer.parseInt(null); Exception in thread "main" java.lang.NumberFormatException: null at java.lang.Integer.parseInt(Integer.java:454) at java.lang.Integer.parseInt(Integer.java:527)
You can see that the error message also contains the argument to parseInt(), which is null, and that’s the reason for this error. Thanks to JDK developers for printing the argument as well and a lesson to all Java developers who create their custom Exception class, always print the input which failed the process. If you want to learn more about best practices, read Java Coding Guidelines.
2. Empty String
Another common reason of java.lang.NumberFormatException is an empty String value. Many developers think empty String is OK and parseInt() will return zero or something, but that’s not true. The parseInt(), or parseFloat() both will throw NumberFormat error as shown below:
Integer.parseInt(""); Exception in thread "main" java.lang.NumberFormatException: For input string: "" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:504) at java.lang.Integer.parseInt(Integer.java:527)
Again you can see that this method has printed the input, which is an empty String and which caused the failure.
3. Alphanumeric Input
Another common reason for NumberFormatException is the alphanumeric input. No non-numeric letter other than + and — is not permitted in the input string.
Short.parseShort("A1"); Exception in thread "main" java.lang.NumberFormatException: For input string: "A1" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:492) at java.lang.Short.parseShort(Short.java:117) at java.lang.Short.parseShort(Short.java:143)
You can see that our input contains «A,» which is not permitted. Hence it caused the error, but you should remember that A is a valid letter when you convert String to an integer in hexadecimal base. See these Java Coding Courses for beginners to learn more about data type conversion in Java.
4. Leading space
You won’t believe but leading, and trailing spaces are one of the major reasons for NumberFormatException in Java; you will think that input String » 123″ is Ok, but it’s not. It has a leading space in it.
Long.parseLong(" 123"); Exception in thread "main" java.lang.NumberFormatException: For input string: " 123" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Long.parseLong(Long.java:430) at java.lang.Long.parseLong(Long.java:483)
Even though the parseLong() method is correctly printing the invalid input, it’s hard to spot the leading or trailing error in log files, particularly the trailing one. So, pay special attention to leading and trailing space while converting String to long in Java; if possible, use trim() before passing String to parseLong() method.
5. Trailing space
Similar to the above issue, trailing space is another main reason for numerous NumberFormatException in Java. A trailing white space is harder to spot than a leading white space, particularly in log files.
Long.parseLong("1001 "); Exception in thread "main" java.lang.NumberFormatException: For input string: "1001 " at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Long.parseLong(Long.java:441) at java.lang.Long.parseLong(Long.java:483)
To avoid this error, you should trim() the input string before passing it to parse methods like the parseInt() or parseFloat().
6. Null String
We’ve already seen a scenario where the parseInt() method throws NumberFormatException if the input is null, but sometimes you will see an error message like Exception in thread «main» java.lang.NumberFormatException: «null» is different than the first scenario.
In the first case, we have a null value; here we have a «null» String i.e., an initialized String object whose value is «null.» Since this is also not numeric, parseInt(), or parseFloat() will throw the NumberFormat exception as shown below.
Float.parseFloat("null"); Exception in thread "main" java.lang.NumberFormatException: For input string: "null" at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:1250) at java.lang.Float.parseFloat(Float.java:452)
If you are not familiar with the concept of null in Java, please see these Java development courses for beginners on Medium.
7. Floating point String
This is a little bit different, even though «1.0» is a perfectly valid String i.e., it doesn’t contain any alphanumeric String, but if you try to convert it into integer values using parseInt(), parseShort(), or parseByte() it will throw NumberFormatException because «1.0» is a floating-point value and cannot be converted into integral one. These methods don’t cast. They just do the conversion.
Long.parseLong("1.0"); Exception in thread "main" java.lang.NumberFormatException: For input string: "1.0" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Long.parseLong(Long.java:441) at java.lang.Long.parseLong(Long.java:483)
So, be careful while converting floating-point String to numbers. They can only be converted to float or double, but not on integral types like byte, short, or int.
8. Out of range value
Another rare reason for java.lang.NumberFormatException is out-of-range value. For example, if you try to convert String «129» to a byte value, it will throw NumberFormatException because the maximum positive number byte can represent is 127. Clearly, 129 is out-of-range for a byte variable
Byte.parseByte("129"); Exception in thread "main" java.lang.NumberFormatException: Value out of range. Value:"129" Radix:10 at java.lang.Byte.parseByte(Byte.java:150) at java.lang.Byte.parseByte(Byte.java:174)
If you are not familiar with data types and their range in Java, please see these free Java programming courses for more information.
9. Non-printable character
While working in the software industry, sometimes you face an issue where you think the computer is wrong, it has gone mad, the input String looks perfectly OK, but still, Java is throwing NumberFormatException, this happens, but at the end, you will realize that computer is always right.
For example, consider this error message:
java.lang.NumberFormatException: For input string: "1" at java.lang.NumberFormatException.forInputString(Unknown Source) at java.lang.Integer.parseInt(Unknown Source) at java.lang.Integer.parseInt(Unknown Source)
«1» is perfectly Ok for parseInt(), and it will run fine then why java.lang.NumberFormatException? It turns out that there was a non-printable character in front of 1, which was not displaying in the place where you are seeing. So, watch out for the non-printable character if you are facing a weird error.
10. Similar looking characters like 1 and l
This is the close cousin of earlier error. This time also our eye believes that input String is valid and thinks that the computer has gone mad, but it wasn’t. Only after spending hours did you realize that Java was right, the String you are trying to convert into a number was not numeric one «1» instead it was a small case letter L, i.e. «l». You can see it’s very subtle and not obvious from the naked eye unless you are really paying attention.
Integer.parseInt("l"); Exception in thread "main" java.lang.NumberFormatException: For input string: "l" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:492) at java.lang.Integer.parseInt(Integer.java:527)
From the error message, it looks like the error is for numeric String «1». So be careful with this.
Here is a nice slide to combine all these reasons into one, which you can share as well :
That’s all about 10 common reasons of java.lang.NumberFormatException and how to solve them. It’s one of those errors where you need to investigate more about data than code. You need to find the source of invalid data and correct it. In code, just make sure you catch the NumberFormatException whenever you convert a string to a number in Java.
Other Java troubleshooting tutorials you may like to explore
- How to fix «Error: Could not find or load main class» in Eclipse? (guide)
- How to avoid ConcurrentModificationException in Java? (tutorial)
- How to solve «could not create the Java virtual machine» error in Java? (solution)
- How to fix «illegal start of expression» compile time error in Java? (tutorial)
- Fixing java.lang.unsupportedclassversionerror unsupported major.minor version 60.0 (solution)
- Cause and solution of «class, interface, or enum expected» compiler error in Java? (fix)
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Also, if you get input from the user, make sure you perform validation. Use Scanner if taking input from the command line and methods like the nextInt() to directly read integer instead of reading String and then parsing it to int again using parseInt() method, only to be greeted by NumberFormatExcpetion.
The NumberFormatException occurs when an attempt is made to convert a string with improper format into a numeric value. That means, when it is not possible to convert a string in any numeric type (float, int, etc), this exception is thrown. It is a Runtime Exception (Unchecked Exception) in Java. It is a subclass of IllegalArgumentException class. To handle this exception, try–catch block can be used.
While operating upon strings, there are times when we need to convert a number represented as a string into an integer type. The method generally used to convert String to Integer in Java is parseInt().
Usage of parseInt() method: As we already know there are two variants of this method namely as follows to get a better understanding
public static int parseInt(String s) throws NumberFormatException This function parses the string argument as a signed decimal integer.
public static int parseInt(String s, int radix) throws NumberFormatException This function parses the string argument as a signed integer in the radix specified by the second argument.
Return Type:
In simple words, both methods convert the string into its integer equivalent. The only difference being is that of the parameter radix. The first method can be considered as an equivalent of the second method with radix = 10 (Decimal).
Constructors:
- public NumberFormatException(): Constructs a NumberFormatException with no detail message.
- public NumberFormatException(String msg): Constructs a NumberFormatException with the detail message ‘msg’
Common Reasons for NumberFormatException:
There are various issues related to improper string format for conversion in numeric value. A few of them are:
1. The input string is null
Integer.parseInt("null") ;
2. The input string is empty
Float.parseFloat(“”) ;
3. The input string with leading and/or trailing white spaces
Integer abc=new Integer(“ 432 “);
4. The input string with extra symbols
Float.parseFloat(4,236);
5. The input string with non-numeric data
Double.parseDouble(“ThirtyFour”);
6. The input string is alphanumeric
Integer.valueOf(“31.two”);
7. The input string might exceed the range of the datatype storing the parsed string
Integer.parseInt(“1326589741236”);
8. Data type mismatch between input string value and the type of the method which is being used for parsing
Integer.parseInt("13.26");
Example:
Java
import java.util.Scanner;
public class GFG {
public static void main(String[] arg)
{
int number;
Scanner sc = new Scanner(System.in);
while (true) {
System.out.println("Enter any valid Integer: ");
try {
number = Integer.parseInt(sc.next());
System.out.println("You entered: "
+ number);
break;
}
catch (NumberFormatException e) {
System.out.println(
"NumberFormatException occurred");
}
}
}
}
Output: The below output is for different numbers been entered by the user
Enter any valid Integer: 12,017 NumberFormatException occurred Enter any valid Integer: Sixty4 NumberFormatException occurred Enter any valid Integer: null NumberFormatException occurred Enter any valid Integer: 436.25 NumberFormatException occurred Enter any valid Integer: 3.o NumberFormatException occurred Enter any valid Integer: 98562341789 NumberFormatException occurred Enter any valid Integer: 1000 You entered: 1000
Last Updated :
18 Feb, 2022
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